Problem
Solve the gold bug story's cipher
Note: This is from Mark Stamp's Information Security: Principles and Practice book, chapter 2. Solution below was submitted as homework for CS265 Cryptography and Information Security class at San Jose State University.
Solution
The cipher
is a substitution cipher. Analysis of the frequency counts of letters and words
in the English language were used to convert the cipher text to plain text.
The original
cipher text is:
53++!305))6*;4826)4+.)4+);806*;48!8`60))85;;]8*;:+*8!83(88)5*!;
46(;88*96*?;8)*+(;485);5*!2:*+(;4956*2(5*-4)8`8*;
4069285);)6
!8)4++;1(+9;48081;8:8+1;48!85;4)485!528806*81(+9;48;(88;4(+?3
4;48)4+;161;:188;+?;
Since e is
the most frequently occurring letter in the English language, and 8 is the most
frequently occurring character (34 times) in the above cipher, we make an
intelligent guess that ‘e’ is mapped to the letter 8. So we get:
53++!305))6*;4e26)4+.)4+);e06*;4e!e`60))e5;;]e*;:+*e!e3(ee)5*!;
46(;ee*96*?;e)*+(;4e5);5*!2:*+(;4956*2(5*-4)e`e*;
40692e5);)6
!e)4++;1(+9;4e0e1;e:e+1;4e!e5;4)4e5!52ee06*e1(+9;4e;(ee;4(+?3
4;4e)4+;161;:1ee;+?;
‘The’ is the
most common 3 word in English with the letter e and correspondingly we find
several instances of “;4e” above. So we guess that ; is t, 4 is h. By replacing
those we get:
53++!305))6*the26)h+.)h+)te06*the!e`60))e5tt]e*t:+*e!e3(ee)5*!t
h6(tee*96*?te)*+(the5)t5*!2:*+(th956*2(5*-h)e`e*t
h0692e5)t)6
!e)h++t1(+9the0e1te:e+1the!e5th)he5!52ee06*e1(+9thet(eeth(+?3
hthe)h+t161t:1eet+?t
We then guess
that the below words in the below order:
t(ee – tree
thr+?3h – through
!egree – degree
th6rtee* - thirteen
5nd – and
|
degree) – degrees
9inutes – minutes
1rom – from
fift: - fifty
0eft – left
|
de`ils – devils
2y – by
bisho.s – bishops
bran-h - branch
t]enty one – twenty one
|
5
|
2
|
-
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!
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8
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1
|
3
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4
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6
|
0
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9
|
*
|
+
|
.
|
(
|
)
|
;
|
?
|
`
|
:
|
]
|
a
|
b
|
c
|
d
|
e
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f
|
g
|
h
|
i
|
l
|
m
|
n
|
o
|
p
|
r
|
s
|
t
|
u
|
v
|
y
|
w
|
Plain text:
a good glass
in the bishops hostel in the devils sea twenty one degrees and thirteen minutes
north east and by north main branch seventh limb east side shoot from the left
eye of the deaths head a beeline from the tree through the shot fifty feet out
This
cryptanalytic success leads to a plain text puzzle with no punctuations. As
Legrand recognizes the words, he adds spaces. Then he splits the words into
groups that make sense, by making use of the fact that people tend overcrowd
details together when they try to make a crypt without punctuations, making the
5 groups stand out.
Groups:
A good glass
in the bishop's hostel in the devil's --twenty-one degrees and thirteen minutes
--northeast and by north --main branch seventh limb east side --shoot from the
left eye of the death's-head --a bee-line from the tree through the shot fifty
feet out.
As a result
of the cryptanalytic success, Ledger figures out the secret location of the
pirate’s (Kidd) treasure from the plain text. With the help of a friend and his
caretaker, he digs out the treasure from the location.
Story - http://poestories.com/read/goldbug